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Whitepunk [10]
3 years ago
6

What is the next term of the geometric sequence 32,16,8

Mathematics
1 answer:
asambeis [7]3 years ago
5 0
32÷2=16
16÷2=8
8÷2=4
4÷2=2
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NikAS [45]

Answer:

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Step-by-step explanation:

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3 0
2 years ago
Which expressions are equivalent to ln x 2 ln 5 ln 1? Check all that apply. 2 ln 5x ln (x 26) ln 25x ln 1 ln 25x.
Marysya12 [62]

Equivalent expression is the expression of two equation when the way of representation of the two equation is different but the result is same.The expressions which are equivalent to the given expression are,

  • \ln25+\ln1
  • \ln25

Given information-

The expression given in the problem is,

\ln x+2 \ln 5 +\ln1

<h3>Equivalent expression-</h3>

Equivalent expression is the expression of two equation when the way of representation of the two equation is different but the result is same.

For given expression,

\ln x+2 \ln 5 +\ln1

As the above function is the function of <em>x. </em>Thus it can be written as,

f(x)=\ln x+2 \ln 5 +\ln1

<h3>Logarithm power rule</h3>

Logarithm power rule states that the exponent of the logarithm function can transfers to front of the logarithm and vice versa. Thus,

f(x)=\ln x+ \ln 5^2 +\ln1\\&#10;f(x)=\ln x+ \ln 25 +\ln1

Now the value of \ln 1 is equal to the zero. Thus,

f(x)=0+\ln 25 +\ln1\\&#10;f(x)=0+\ln 25 +\ln1\\&#10;f(x)=\ln 25 +\ln1\\&#10;f(x)=\ln 25 +0\\&#10;f(x)=\ln 25

Hence the expressions which are equivalent to the given expression are,

  • \ln25+\ln1
  • \ln25

Learn more about the equivalent expression here;

brainly.com/question/10628562

5 0
2 years ago
Read 2 more answers
21. In 4 + In(4x - 15) = In(5x + 19)
Alexxx [7]

Answer:

x=-30;\quad \:I\ne \:0

Step-by-step explanation:

In\cdot \:4+In\left(4x-15\right)=In\left(5x+19\right)

\:4+In\left(4x-15\right):\quad -11nI+4nxI\\In\cdot \:4+In\left(4x-15\right)\\=4nI+nI\left(4x-15\right)\\\\\:In\left(4x-15\right):\quad 4nxI-15nI\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b-c\right)=ab-ac\\a=In,\:b=4x,\:c=15\\=In\cdot \:4x-In\cdot \:15\\=4nxI-15nI\\=In\cdot \:4+4nxI-15nI\\\mathrm{Simplify}\:In\cdot \:4+4nxI-15nI:\quad -11nI+4nxI\\In\cdot \:4+4nxI-15nI\\\mathrm{Group\:like\:terms}\\=4nI-15nI+4nxI\\\mathrm{Add\:similar\:elements:}\:4nI-15nI=-11nI\\=-11nI+4nxI\\

\mathrm{Expand\:}In\left(5x+19\right):\quad 5nxI+19nI\\In\left(5x+19\right)\\=nI\left(5x+19\right)\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac\\a=In,\:b=5x,\:c=19\\=In\cdot \:5x+In\cdot \:19\\=5nxI+19nI\\\\-11nI+4nxI=5nxI+19nI\\\\\mathrm{Add\:}11nI\mathrm{\:to\:both\:sides}\\-11nI+4nxI+11nI=5nxI+19nI+11nI\\Simplify\\4nxI=5nxI+30nI\\\mathrm{Subtract\:}5nxI\mathrm{\:from\:both\:sides}\\4nxI-5nxI=5nxI+30nI-5nxI\\\mathrm{Simplify}\\-nxI=30nI\\

\mathrm{Divide\:both\:sides\:by\:}-nI;\quad \:I\ne \:0\\\frac{-nxI}{-nI}=\frac{30nI}{-nI};\quad \:I\ne \:0\\\mathrm{Simplify}\\\frac{-nxI}{-nI}=\frac{30nI}{-nI}\\\mathrm{Simplify\:}\frac{-nxI}{-nI}:\quad x\\\frac{-nxI}{-nI}\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{-b}=\frac{a}{b}\\=\frac{nxI}{nI}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{xI}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=x\\\mathrm{Simplify\:}\frac{30nI}{-nI}:\quad -30\\\mathrm{Apply\:the\:fraction\:rule}:

\quad \frac{a}{-b}=-\frac{a}{b}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{30I}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=-30\\x=-30;\quad \:I\ne \:0

3 0
3 years ago
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m
borishaifa [10]

Answer:

the concentration at 10 minutes= 0.4+0.0133= 0.4133%

Step-by-step explanation:

Air containing 0.04% carbon dioxide

V, volume of room is 6000 ft3.

Q, rate of air 2000 ft3/min,

initial concentration of 0.4% carbon dioxide,

determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes?

firstly, we find the time taken for air to completely filled the room

Q = V/t

t = V/Q = 6000/2000 = 3min

so, its take 3mins for air to be completely filled in the room and for exhaust air to move out.

there is  an initial concentration of 0.4% carbon dioxide, and the air pump in is 0.04%.

therefore,

3mins = 0.04% of CO2

3*60 =180sec = 0.04%

1sec = 0.04/180 = 0.00022%/sec

so at any time the concentration of CO2 is 0.4 + 0.00022 =0.40022%/sec

What is the concentration at 10 minute

the concentration at 10minutes = the concentration for 1minute because at every minutes, the concentration moves in is moves out. = concentration for 2000ft3.

for 0.04% = 6000ft3

   ?          = 2000ft3

              = 2000* 0.04)/6000 =0.0133%

the concentration at 10 minutes= 0.4+0.0133= 0.4133%

4 0
3 years ago
Can someone help me out with this ? 8(1 - 8p) + 8p
goldfiish [28.3K]
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6 0
2 years ago
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